mathop {lim }limits_{x to 0} frac{{1 - cos x} | vedclass

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(A) We evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{1 - \cos x}}{x}$.Using the trigonometric identity $1 - \cos x = 2 \sin^2(x/2)$,the

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(A) We evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{1 - \cos x}}{x}$.Using the trigonometric identity $1 - \cos x = 2 \sin^2(x/2)$,the expression becomes:$\mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2(x/2)}{x}$.Multiply and divide by $x/2$ to use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$:$\mathop {\lim }\limits_{x 	o 0} \left( \frac{2 \sin^2(x/2)}{x} \cdot \frac{x/4}{x/4} ight) = \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \left( \frac{\sin(x/2)}{x/2} ight)^2 \cdot \frac{x}{4} ight) = 2 \cdot (1)^2 \cdot 0 = 0$.Alternatively,using $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(1 - \cos x)}{\frac{d}{dx}(x)} = \mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{1} = \sin(0) = 0$.

$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{x} = $

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